An asphalt pavement drains into a rectangular concrete channel. the catchment surface has an average slope of 1.0%, and the distance from the catchment boundary to the drain is 30 m. the drainage channel is 60 m long, 20 cm wide, and 25 cm deep, and has a slope of 0.6%. for an effective rainfall rate of 50 mm/h, the flowrate in the channel is estimated to be 0.02 m3 /s. the manning n for an asphalt surface is 0.011 and for a concrete surface is 0.013. estimate the time of concentration of overland flow for the catchment. (hint: use the kinematic-wave equation for overland flow.)

Step-by-step explanation:

Using kinematic-wave equation to calculate the time of concentration

Where is time of concentration in minutes, n is Manning’s roughness coefficient, L is travel distance between catchment boundary and drain in metres, is effective rainfall intensity in mm/h, is the slope of the ground

Substituting L with 30 m as given, with 50 mm/h as provided, with 0.01 as given, Manning’s coefficient with 0.011 as given then  

The area, A of rectangular channel in terms of channel depth

A=wd where d is depth and w is channel width

Substituting w with 20cm which is 0.2m

A=0.2d

Wetted perimeter of rectangular channel

P=2d+w=20d+0.2

From Manning’s equation of flow

where q is flow rate, A is cross-sectionall area, R is hydraulic radius, is the slope and n is Manning’s coefficient

The flow rate, q is given in the question as 0.02 m3/s and Manning’s coefficient for overland flow which corresponds to concrete is given as 0.013

d=0.12m

We can now get the flow area which we already stated is 0.2d and now we have d

To find the velocity of flow in the drainage channel, we divide discharge by velocity hence

The time of concentration in the catchment area,

Total time of concentration,

Therefore,

horizontal translation of 2 units in the positive x direction

followed by

vertical translation of 3 units in the positive y direction

1,150 give brainliest

step-by-step explanation

2r= 2300

r =1150